如何在PHP（多平台）中获取给定周数的第一天？

Question

What is the simplest way to do it in PHP ?

I want the date of the Monday of a given week number of a year (example : week number 3 of 2009)

Thanks !

EDIT : If you use Linux only machines, use cletus' solution, however I am looking for something that can work on Windows AND Linux.

Answer 1

It's simple on PHP 5.3

echo date('M d',strtotime('2013W15'));

where 15 is the number of week. But for the number below ten make sure it is in the format of 01, 02 for first week and second week.

Answer 2

Yet another solution:

<?php
        $week = 3;
        $year = 2009;

        $timestamp = mktime( 0, 0, 0, 1, 1,  $year ) + ( $week * 7 * 24 * 60 * 60 );
        $timestamp_for_monday = $timestamp - 86400 * ( date( 'N', $timestamp ) - 1 );
        $date_for_monday = date( 'Y-m-d', $timestamp_for_monday );
?>

Answer 3

A nice way to get this in a clean way is by using php DateTime class.

$year = 2015;
$week_no = 1;

$date = new DateTime();
$date->setISODate($year,$week_no);
echo $date->format('d-M-Y'); 

This would result into : 29-12-2014

Answer 4

You can use strptime() to get the time.

$time = strptime('1 23 2009', '%w %U %Y');

This will get the time for the Monday (day 1, 0 is Sunday, 6 is Saturday) of the 23rd week of 2009. If you want to format this into a date, use date() .

$date = date('d F Y', $time);

Answer 5

Seems to be working and not dependent of the server OS :

<?php

  $week = 4;
  $year = 2013;

  $timestamp_for_monday = mktime( 0, 0, 0, 1, 1,  $year ) + ((7+1-(date( 'N', mktime( 0, 0, 0, 1, 1,  $year ) )))*86400) + ($week-2)*7*86400 + 1 ;

?>

the idea is to add :

• the timestamp of the first of January of the chosen year
• the number of seconds to reach the end of the first week (which is 7 days minus the day of week of the 1st of January + 1 day) multiplied by the number of seconds per day
• the number of seconds of the number of chosen weeks minus the first week and the current week
• 1 second to reach the fist second of the current week

My example returns : 1358722801 which is the timestamp of 2013/01/21 0:00:01

Answer 6

This next script gives the 7 days of an specific week of a year

$time = new DateTime();
$time->setISODate(2016, 13);
for($i=0;$i<7;$i++){
    echo $time->format('d-M-Y') . '<br />';
    $time->add(new DateInterval('P1D'));
}

Answer 7

Here is very simple solution, passing a week no and returns the date.

The ISO8601 standard states that week 1 always fall on the week where Jan 4 falls.

For example, to get a day in the 4th week of the year:

$day_in_week = strtotime("2006-01-04 + 4 weeks"));

Then you can adjust this value to Sunday (as a starting place you can guarantee that you can find):

// Find that day's day of the week (value of 0-6)
$wday = date('w', $day_in_week);
$offset = 6 - $wday; // How far it is from Sunday.
$sunday_in_week = $day_in_week - ($offset * (60 * 60 * 24)); // $offset * seconds in a day

Then, you add the seconds in a day again to get Monday.

$monday_in_week = $sunday_in_week + (60 * 60 * 24);

Note: This method can occasionally have some problems with daylight savings time. A similar, and slightly safer between DST time changes, method would use the DateTime class . However, DateTime is only support in PHP 5.2.0 or later. The method above works in earlier version as well.

Answer 8

Try this function

function MondayOfWeek($WeekNumber, $Year=-1) {
    if ($Year == -1) $Year   = 0+date("Y");
    $NewYearDate             = mktime(0,0,0,1,1,$Year);
    $FirstMondayDate         = 7 + 1 - date("w", mktime(0,0,0,1,1,2009));
    $Dates_fromFirstMonday   = 7 * $WeekNumber;
    $Second_fromFirstMonday  = 60*60*24*($FirstMondayDate + $Dates_fromFirstMonday);
    $MondayDay_ofWeek        = $NewYearDate + $Second_fromFirstMonday;
    $Date_ofMondayDay_ofWeek = 0+date("j", $MondayDay_ofWeek);
    return $Date_ofMondayDay_ofWeek;
}

for($i = 0; $i 

When run, I got:

-5-12-19-26-2-9-16-23-2-9-16-23-30-6-13-20-27-4-11-18-25-1-8-15-22-29-6-13-20-27-3-10-17-24-31-7-14-21-28-5-12-19-26-2-9-16-23-30-7-14-21-28

Hope this helps.

Answer 9

I required same in the java script..so i converted.

function(week,year){
var timestamp = new Date(year, 0, 1, 0, 0, 0, 0);
var dateObj=new Date();
var val = timestamp.getTime(); 
days=( week * 7 * 24 * 60 * 60*1000 );
val=val+days;
var timestamp_for_monday =val - 86400 *((timestamp.getDay()-1000));
var weekdate=new Date(timestamp_for_monday);
return weekdate;
}